5.2: The Laplace Transform - Mathematics

The Laplace Transform is typically credited with taking dynamical problems into static problems. Recall that the Laplace Transform of the function (h) is

[mathscr{L} (h(s)) equiv int_{0}^{infty} e^{-(st)} h(t) dt onumber]

MATLAB is very adept at such things. For example:

The Laplace Transform in MATLAB

>> syms t>> laplace(exp(t))ans = 1/(s-1)>> laplace(t*(exp(-t))ans = 1/(s+1)^2

The Laplace Transform of a matrix of functions is simply the matrix of Laplace transforms of the individual elements.

Definition: Laplace Transform of a matrix of fucntions

[mathscr{L} (egin{pmatrix} {e^{t}} {te^{-t}} end{pmatrix}) = egin{pmatrix} {frac{1}{s-1}} {frac{1}{(s+1)^2}} end{pmatrix} onumber]

Now, in preparing to apply the Laplace transform to our equation from the dynamic strang quartet module:

[ extbf{x}' = B extbf{x}+ extbf{g}]

we write it as

[mathscr{L} (frac{dx}{dt}) = mathscr{L}(B extbf{x}+ extbf{g})]

and so must determine how (mathscr{L}) acts on derivatives and sums. With respect to the latter it follows directly from the definition that

[ egin{align*} mathscr{L}(B extbf{x}+ extbf{g}) &= mathscr{L}(B extbf{x})+mathscr{L}( extbf{g}) [4pt] &= B mathscr{L}( extbf{x})+mathscr{L}( extbf{g}) end{align*}]

Regarding its effect on the derivative we find, on integrating by parts, that

[egin{align} mathscr{L} left(frac{d extbf{x}}{dt} ight) &= int_{0}^{infty} e^{-(st)} frac{d extbf{x}(t)}{dt} dt [4pt] &= extbf{x}(t) left. e^{-(st)} ight|_{0}^{infty}+sint_{0}^{infty} e^{-(st)} extbf{x}(t) dt end{align}]

Supposing that (x) and (s) are such that (x(t) e^{-(st)} ightarrow 0) as (t ightarrow infty) we arrive at

[mathscr{L} (frac{d extbf{x}}{dt}) = smathscr{L} ( extbf{x})-x(0) onumber]

Now, upon substituting Equation 2 and Equation 3 into Equation 1 we find

[s mathscr{L} ( extbf{x})- extbf{x}(0) = B mathscr{L}( extbf{x})+mathscr{L}( extbf{g}) onumber]

which is easily recognized to be a linear system for (mathscr{L}( extbf{x}))

[( extbf{s}I-B) mathscr{L}( extbf{x}) = mathscr{L}( extbf{g})+x(0) onumber]

The only thing that distinguishes this system from those encountered since our first brush with these systems is the presence of the complex variable (s). This complicates the mechanical steps of Gaussian Elimination or the Gauss-Jordan Method but the methods indeed apply without change. Taking up the latter method, we write

[mathscr{L}( extbf{x}) = (sI-B)^{-1} (mathscr{L}( extbf{g})+x(0)) onumber]

The matrix ((sI-B)^{-1}) is typically called the transfer function or resolvent, associated with (B), at (s). We turn to MATLAB for its symbolic calculation. (for more information, see the tutorial on MATLAB's symbolic toolbox). For example,

>> B = [2 -1; -1 2]>> R = inv(s*eye(2)-B)R =[ (s-2)/(s*s-4*s+3), -1/(s*s-4*s+3)][ -1/(s*s-4*s+3), (s-2)/(s*s-4*s+3)]

We note that ((sI-B)^{-1}) well defined except at the roots of the quadratic, (s^{2}-4s+3) determinant of ((sI-B)) and is often referred to as the characteristic polynomial of (B). Its roots are called the eigenvalues of (B).

Example (PageIndex{1})

Let us take the (B) matrix of the dynamic Strang quartet module with the parameter choices specified in fib3.m, namely

[B = egin{pmatrix} {-0.135}&{0.125}&{0} {0.5}&{-1.01}&{0.5} {0}&{0.5}&{-0.51} end{pmatrix} onumber]

The associated ((sI-B)^{-1}) is a bit bulky (please run fig3.m) so we display here only the denominator of each term, i.e.,

[s^3+1.655s^2+0.4078s+0.0039 onumber]

Assuming a current stimulus of the form (i_{0}(t) = frac{t^{3}e^{-frac{t}{6}}}{10000}) and (E_{m} = 0) brings

[mathscr{L}( extbf{g})(s) = egin{pmatrix} {frac{0.191}{(s+frac{1}{6})^{4}}} {0} {0} {0} end{pmatrix} onumber]

and so Equation persists in

[egin{align*} mathscr{L}( extbf{x}) &= (sI-B)^{-1} mathscr{L}( extbf{g}) [4pt] &= frac{0.191}{(s+frac{1}{6})^{4}(s^3+1.655s^2+0.4078s+0.0039)} egin{pmatrix} {s^2+1.5s+0.27} {0.5s+0.26} {0.2497} end{pmatrix} end{align*}]

Now comes the rub. A simple linear solve (or inversion) has left us with the Laplace transform of ( extbf{x}). The accursed No Free Lunch Theorem

We shall have to do some work in order to recover ( extbf{x}) from (mathscr{L}( extbf{x})) confronts us. We shall face it down in the Inverse Laplace module.

The Laplace Transform

The Laplace Transform applied to a function $f(t)$ will look like the following:
$mathcal[f(t)]=int_0^ e^ <-lambda t>f(t) ,dt = F(lambda)$
Shorthand notation is usually denoted in 3 common ways:
$mathcal[f(t)] = F^ = Laplace(f(t))$
In general, your function of time $f(t)$ will be transformed into a function of lambda $F(lambda)$:
$f(t)$ =>$F^(lambda)$

Please note the following properties of the Laplace Transform:
Always remember that the Laplace Transform is only valid for t>0.
Constants can be pulled out of the Laplace Transform:
$mathcal[af(t)] = amathcal[f(t)]$ where a is a constant
Also, the Laplace of a sum of multiple functions can be split up into the sum of multiple Laplace Transforms:
$mathcal[g(t)+f(t)] = mathcal[g(t)]+mathcal[f(t)]$

There are 5 rules that you should memorize about the Laplace Transform:

1. Convolution Rule
We will denote the convolution of 2 functions f and g as the following:
$(f * g) = (g * f) = int_<0>^ f( au) g(t- au)mathrm au$
When we apply the Laplace Transform to the convolution of 2 functions we obtain the following result:
$mathcal[f * g] = F^(lambda)G^(lambda)$

2. Derivative Rule
Given a derivative (n) of a function f, denoted by $f^$, the Laplace Transform will be the following:
$mathcal[f^] = lambda^F^(lambda)-sum_^lambda^f^(+0)$

3. Similarity Rule
Given a function that has a constant $a$ multiplied by t in a function:
$mathcal[f(at)] = frac<1>F^(frac),$ such that $a>0$

4. Shift Rule
Given a function shifted by a certain amount multiplied by a shifted Heaviside Function:
$mathcal[H(t-a)g(t-a))] = e^<-alambda>G^(lambda)$

5. Attenuation rule
Given an exponential function multiplied by an exponential function, where a is a constant:
$mathcal[e^<-at>f(t)] = F^(lambda+a)$

Note that the most important rules that we will use are #1, #2, and #4, however it is a good idea to learn all of the rules.

Module-I(T-3 h + Pj-2 h)

Laplace Transforms, Properties of Laplace transforms, Unit step function.

Make a short draft of properties of Laplace transform from memory. Then compare your notes with the text and write a report of 2-3 pages on these operations and their significance in applications.

Module-II(T-2 h + Pj-2 h)

Second shifting theorem, Laplace transforms of Derivatives and Integrals.

Find the Laplace transform of the following functions.

Module-III(T-3 h + Pj-2 h)

Derivatives and Integrals of Transforms, Inverse Laplace transform.

Application of Unit step function (RC- Circuit to a single square wave).

Module-IV(T-2 h + Pj-2 h)

Solution of differential equations by using Laplace transform.

Find the solution of differential equation by using Laplace Transform.

Module-V(T-4 h + Pj-2 h)

Periodic function, Fourier series, Fourier series expansion of an arbitrary period, Half range expansions.

Find the Fourier series expansion of a 2-pi periodic function.

Module-VI(T-3 h + Pj-2 h)

Complex form of Fourier series, Fourier Integrals, Different forms of Fourier Integral.

Laplace Transforms

§8.6 Historical Epilogue

Oliver Heaviside’s legacy to mathematics and electromagnetism is impressive. In addition to perfecting the operational calculus that later inspired the Laplace transform method , he developed vector calculus in 1885, starting with the definitions of scalar and vector products as used today ( EPII, pages 4 and 5 ). 1 In the same year he formulated what has become the cornerstone of electromagnetic theory. Heaviside refers to his discovery as follows:

I here introduce a new method of treating the subject [Maxwell’s theory of electromagnetism], which may perhaps be appropriately named the Duplex method, since its main characteristic is the exhibition of the electric, magnetic, and electromagnetic equations in a duplex form.

This was the first appearance in print of the famous Maxwell’s equations of electromagnetic theory (EPI, pages 447 , 448 , 452 and 475 ), which are not included in Maxwell’s treatise. Maxwell was not a very clear writer—his treatise is almost unreadable from a certain point on—so that many future expositors of the subject preferred to follow Heaviside’s interpretation of Maxwell and did not realize that the duplex equations were Heaviside’s own.

From Electrical Papers, Vol. 1, The Copley Publishers, Boston, 1925.

Heaviside’s contribution to telegraphy and telephony was invaluable, but for the longest time they fell on deaf ears in his own country. He found a formidable obstacle in William Henry Preece, Electrician to the Post Office. Preece’s opposition was based on a two-pillar foundation. The first was his ignorance of what really took place in the transmission of electric signals, as shown in a paper published in 1887. The second was Heaviside’s reply containing the following assessment:

Either, firstly, the accepted theory of electromagnetism must be most profoundly modified or, secondly, the views expressed by Mr. Preece in his paper are profoundly erroneous … Mr. Preece is wrong, not merely in some points of detail, but radically wrong, generally speaking, in methods, reasoning, results, and conclusions.

The fact is that Mr. Preece, later Sir William, never accepted the validity of Heav-iside’s advice to increase both the inductance and, to a lesser extent to avoid excessive attenuation, the leakage conductance, so as to approximate the distortionless condition g/c = r/l. "More capacitance," seems to have been Preece’s motto. The outcome of all this is that Heaviside started having trouble publishing his papers, which were in opposition to the official view, that the British government sank a fortune building the wrong kind of lines, and that England lost its leadership in this field to America. Mihajlo Idvorsky Pupin, a Serbian immigrant from the Austrian village of Idvor who became professor of mathematics at Columbia University, was the first to build a line using increased inductance, but in his paper of May 1900 acknowledged the source of his theoretical background, stating that

Mr. Oliver Heaviside of England, to whose profound researches most of the existing mathematical theory of electrical wave propagation is due, was the originator and most ardent advocate of wave conductors of high inductance.

Soon afterwords, the American Telephone and Telegraph Company succeeded in establishing coast to coast telephone communication by using increased inductance.

Heaviside was good with words in many ways. To him we owe, for instance, the terms inductance, attenuation, and magnetic reluctance (EPII, page 28 ) and the use of voltage for electromotive force (EMTI, page 26 ). He was a colorful, entertaining and opinionated writer, as shown by the following additional quotations:

Self-induction is salvation. [EMTII, page 354 ] As critics cannot always find time to read more than the preface, the following remarks may serve to direct their attention to some of the leading points in this volume. [EMTI, Preface] And that there is a natural tendency for both the human body and understanding to move in circles is proved by the accounts of the doings of belated travellers in the wild, and by the contents of a great mass of books. [EPI, page 353 ] Electric and magnetic force. May they live for ever and never be forgot. [EMTIII, page 1 ] When Prof. Hughes speaks of the resistance of a wire, he does not always mean what common men, men of ohms, volts and farads, mean by the resistance of a wire—only sometimes. [EPII, page 28 ] Different men have different opinions—some like apples, some like inions [EMTI, page 352 ].

Mathematician at large, electrician, philosopher, acid humorist, iconoclast extraordinaire, he was awarded—but declined—the Hughes Medal of the Royal Society in 1904, received a Ph.D honoris causa from the University of Göttingen in 1906, was made an honorary member of the Institution of Electrical Engineers of Great Britain in 1908 and of the American Institute of Electrical Engineers in 1918, and was awarded the first Faraday Medal of the Institution of Electrical Engineers in 1923.

At his country home in Torquay, where he spent the last seventeen years of his life mostly alone and in great financial trouble—despite a small government pension that he accepted only on the condition that it be in recognition of his scientific work—things were less rosy. For lack of payment the bank was after his home, and the gas company cut off his gas. A victim of lumbago and rheumatic gout, he had to eat cold food and live in a cold house. On arriving at his door in the winter of 1921, a distinguished visitor found a note stating that Heaviside had gone to bed to keep warm. Stuffed in the cracks of the door, to prevent any cold drafts, there was an assortment of papers: some advertisements, an invitation by the President of the Royal Society, threats from the gas company about cutting off the gas …. The following Spring Heaviside wrote:

Could not wear boots at all. Could not get proper bed socks to walk about in. Buried under all the blankets I have. Now and then I scribbled a sort of diary about my persecution by the Poor and the Gas and others. 1

Irrepressible in his writing, he continued working on his scientific papers, many of which were found posthumously. He died in a nursing home on February 3, 1925.

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Discrete Systems Properties of Fourier transforms

Many properties of Laplace and Fourier transforms are quite similar. In particular, most of the formulas collected in Appendix 7 can be adapted to Fourier transforms by using the rule of correspondence [8.18] . In particular, the convolution theorem [A7.8] can be extended to Fourier transforms, provided the lower limit of integration is properly shifted from 0 to - ∞. Indeed, a major difference between the two transforms is that in the Fourier integral, time origin does not play any particular role, in contrast to the case of the Laplace integral. As a further consequence of particular interest in dynamics, the terms connected to the initial values in the differentiation theorem [A7.3] , have to be dropped when adapting the formula to the case of the Fourier transforms. In consequence, motion calculated by using a Fourier transformation of the dynamical equations automatically discards the free oscillations induced by nonzero initial conditions of the external excitation. This is a suitable property when interest is restricted to the study of the steady regime of forced responses.

250+ TOP MCQs on The Laplace Transform and Answers

Signals & Systems Multiple Choice Questions on “The Laplace Transform”.

1. The necessary condition for convergence of the Laplace transform is the absolute integrability of f(t)e -σt .
A. True
B. False
Answer: A
Clarification: The necessary condition for convergence of the Laplace transform is the absolute integrability of f(t)e -σt .Mathematically, this can be stated as
(int_<-∞>^∞|f(t) e^<-σt>|)dt -at u(t) and its ROC.
A. (frac<1>), Re>-a
B. (frac<1>), Re>a
C. (frac<1>), Re>a
D. (frac<1>), Re>-a
Answer: D
Clarification: Laplace transform, L = X(s) = (int_<-∞>^∞ x(t) e^ <-st>,dt)
L = (int_<-∞>^∞ e^ <-at>u(t) e^ <-st>,dt = int_0^∞ e^ <-at>e^ <-st>,dt = frac<1>) when (s+A.>0
ROC is Re>-a.

3. Find the Laplace transform of δ(t).
A. 1
B. 0
C. ∞
D. 2
Answer: A
Clarification: Laplace transform, L = X(s) = (int_<-∞>^∞ x(t) e^ <-st>,dt)
L <δ(t)>= (int_<-∞>^∞ δ(t) e^ <-st>,dt)
[x(t)δ(t) = x(0)δ(t)]
= (int_<-∞>^∞ δ(t)dt)
= 1.

4. Find the Laplace transform of u(t) and its ROC.
A. (frac<1>), σ 0
C. (frac<1>), σ=0
D. (frac<1><1-s>), σ≤0
Answer: B
Clarification: Laplace transform, L = X(s) = (int_<-∞>^∞ x(t) e^ <-st>,dt)
L = (int_<-∞>^∞ u(t) e^ <-st>,dt = int_0^∞ e^ <-st>,dt = frac<1>) when s>0 i.e,σ>0.

5. Find the ROC of x(t) = e -2t u(t) + e -3t u(t).
A. σ>2
B. σ>3
C. σ>-3
D. σ>-2
Answer: D
Clarification: Given x(t) = e -2t u(t) + e -3t u(t)
Laplace transform, L = X(s) = (int_<-∞>^∞ x(t) e^ <-st>,dt)
X(s) = (frac<1> + frac<1>)
ROC is <σ>-2>∩ <σ>-3>
Hence, the ROC is σ > -2.

How Laplace Transform Calculator Works?

An online Laplace transformation calculator helps you to transform real functions into complex function with these steps:


  • First, enter a simple equation, and you can see the equation preview.
  • Hit the calculate button for further process.


The Laplace transform calculator displays the following results:

  • First of all, the calculator shows your input in the form of the ordinary differential equation.
  • Then, provide the answer against the equation in algebraic form.

5.2: The Laplace Transform - Mathematics

The Laplace transform is an Integral Transform perhaps second only to the Fourier Transform in its utility in solving physical problems. Due to its useful properties, the Laplace transform is particularly useful in solving linear Ordinary Differential Equations such as those arising in the analysis of electronic circuits.

The (one-sided) Laplace transform (not to be confused with the Lie Derivative) is defined by

where is defined for . A two-sided Laplace transform is sometimes also defined by

The Laplace transform existence theorem states that, if is piecewise Continuous on every finite interval in satisfying

for all , then exists for all . The Laplace transform is also Unique, in the sense that, given two functions and with the same transform so that

then Lerch's Theorem guarantees that the integral

vanishes for all for a Null Function defined by

In the above table, is the zeroth order Bessel Function of the First Kind, is the Delta Function, and is the Heaviside Step Function. The Laplace transform has many important properties.

The Laplace transform of a Convolution is given by

Now consider Differentiation. Let be continuously differentiable times in . If , then

Continuing for higher order derivatives then gives

This property can be used to transform differential equations into algebraic equations, a procedure known as the Heaviside Calculus, which can then be inverse transformed to obtain the solution. For example, applying the Laplace transform to the equation

which can be rearranged to

If this equation can be inverse Laplace transformed, then the original differential equation is solved.

Consider Exponentiation. If for , then for .

Consider Integration. If is piecewise continuous and , then

The inverse transform is known as the Bromwich Integral, or sometimes the Fourier-Mellin Integral.

Arfken, G. Mathematical Methods for Physicists, 3rd ed. Orlando, FL: Academic Press, pp. 824-863, 1985.

Churchill, R. V. Operational Mathematics. New York: McGraw-Hill, 1958.

Franklin, P. An Introduction to Fourier Methods and the Laplace Transformation. New York: Dover, 1958.

Morse, P. M. and Feshbach, H. Methods of Theoretical Physics, Part I. New York: McGraw-Hill, pp. 467-469, 1953.

Spiegel, M. R. Theory and Problems of Laplace Transforms. New York: McGraw-Hill, 1965.

Widder, D. V. The Laplace Transform. Princeton, NJ: Princeton University Press, 1941.


The Laplace transform is defined as a unilateral or one-sided transform. This definition assumes that the signal f(t) is only defined for all real numbers t ≥ 0 , or f(t) = 0 for t < 0 . Therefore, for a generalized signal with f(t) ≠ 0 for t < 0 , the Laplace transform of f(t) gives the same result as if f(t) is multiplied by a Heaviside step function.

Watch the video: Laplace Transform Pairs (October 2021).